If $a, b, c, d$ are in proportion, prove that :

$\begin{array}{ll} \text{(i)} & (5a + 7b)(2c - 3d) \ & = (5c + 7d)(2a - 3b) \\ \text{(ii)} & (ma + nb) : b \ & = (mc + nd) : d \\ \text{(iii)} & (a^4 + c^4) : (b^4 + d^4) \ & = a^2c^2 : b^2d^2 \\ \text{(iv)} & \dfrac{a^2 + ab}{c^2 + cd} = \dfrac{b^2 - 2ab}{d^2 - 2cd} \\[1em] \text{(v)} & \dfrac{(a + c)^3}{(b + d)^3} = \dfrac{a(a - c)^2}{b(b - d)^2} \\[1em] \text{(vi)} & \dfrac{a^2 + ab + b^2}{a^2 - ab + b^2} \ & = \dfrac{c^2 + cd + d^2}{c^2 - cd + d^2} \\ \text{(vii)} & \dfrac{a^2 + b^2}{c^2 + d^2} = \dfrac{ab + ad - bc}{bc + cd - ad} \\[1em] \text{(viii)} & abcd\Big(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2} \Big) \ & = a^2 + b^2 + c^2 + d^2. \end{array}$

If x, y, z are in continued proportion, prove that : $\dfrac{(x + y)^2}{(y + z)^2} = \dfrac{x}{z}.$

If $a, b, c$ are in continued proportion prove that :

$\begin{array}{ll} \text{(i)} & \dfrac{a + b}{b + c} = \dfrac{a^2(b - c)}{b^2(a - b)} \\ \text{(ii)} & \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} \ & = \dfrac{a}{b^2c^2} + \dfrac{b}{c^2a^2} + \dfrac{c}{a^2b^2} \\ \text{(iii)} & a : c = (a^2 + b^2) : (b^2 + c^2) \\ \text{(iv)} & a^2b^2c^2(a^{-4} + b^{-4} + c^{-4}) \ & = b^{-2}(a^4 + b^4 + c^4) \\ \text{(v)} & abc(a + b + c)^3 \ & = (ab + bc + ca)^3 \\ \text{(vi)} & (a + b + c)(a - b + c) \ & = a^2 + b^2 + c^2. \end{array}$

If a, b, c, d are in continued proportion, prove that :

(i) $\dfrac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \dfrac{a}{d}$

(ii) (a² - b²)(c² - d²) = (b² - c²)²

(iii) (a + d)(b + c) - (a + c)(b + d) = (b - c)²

(iv) a : d = triplicate ratio of (a - b) : (b - c)

(v) $\big(\dfrac{a - b}{c} + \dfrac{a - c}{b}\big)^2 - \big(\dfrac{d - b}{c} + \dfrac{d - c}{b}\big)^2 = (a - d)^2\big(\dfrac{1}{c^2} - \dfrac{1}{b^2}\big).$